Again this theorem is too difficult to prove here, but a special case is easier. In the proof of a special case of Green's Theorem, we needed to know that we could describe the region of integration in both possible orders, so that we could set up one double integral using \(dx\,dy\) and another using \(dy\,dx\). Similarly here, we need to be able to describe the three-dimensional region \(E\) in different ways.

where \(B\) is the region in the \(y\)-\(z\) plane over which we integrate. The boundary surface of \(E\) consists of a "top'' \(x=g_2(y,z)\), a "bottom'' \(x=g_1(y,z)\), and a "wrap-around side'' that is vertical to the \(y\)-\(z\) plane. To integrate over the entire boundary surface, we can integrate over each of these (top, bottom, side) and add the results. Over the side surface, the vector \(\bf N\) is perpendicular to the vector \(\bf i\), so

Thus, we are left with just the surface integral over the top plus the surface integral over the bottom. For the top, we use the vector function \({\bf r}=\langle g_2(y,z),y,z\rangle\) which gives \({\bf r}_y\times{\bf r}_z=\langle 1,-g_{2y},-g_{2z}\rangle\); the dot product of this with \({\bf i}=\langle 1,0,0\rangle\) is 1. Then

Let \({\bf F}=\langle 2x,3y,z^2\rangle\), and consider the three-dimensional volume inside the cube with faces parallel to the principal planes and opposite corners at \((0,0,0)\) and \((1,1,1)\). We compute the two integrals of the divergence theorem.

The triple integral is the easier of the two:

$$\int_0^1\int_0^1\int_0^1 2+3+2z\,dx\,dy\,dz=6.$$

The surface integral must be separated into six parts, one for each face of the cube. One face is \(z=0\) or \({\bf r}=\langle u,v,0\rangle\), \(0\le u,v\le 1\). Then \({\bf r}_u=\langle 1,0,0\rangle\), \({\bf r}_v=\langle 0,1,0\rangle\), and \({\bf r}_u\times{\bf r}_v= \langle 0,0,1\rangle\). We need this to be oriented downward (out of the cube), so we use \(\langle 0,0,-1\rangle\) and the corresponding integral is

$$\int_0^1\int_0^1 -z^2\,du\,dv=\int_0^1\int_0^1 0\,du\,dv=0.$$

Another face is \(y=1\) or \({\bf r}=\langle u,1,v\rangle\). Then \({\bf r}_u=\langle 1,0,0\rangle\), \({\bf r}_v=\langle 0,0,1\rangle\), and \({\bf r}_u\times{\bf r}_v= \langle 0,-1,0\rangle\). We need a normal in the positive \(y\) direction, so we convert this to \(\langle 0,1,0\rangle\), and the corresponding integral is

$$\int_0^1\int_0^1 3y\,du\,dv=\int_0^1\int_0^1 3\,du\,dv=3.$$

The remaining four integrals have values 0, 0, 2, and 1, and the sum of these is 6, in agreement with the triple integral.

For the surface we need three integrals. The top of the cylinder can be represented by \({\bf r}=\langle v\cos u,v\sin u,2\rangle\); \({\bf r}_u\times{\bf r}_v=\langle 0,0,-v\rangle\), which points down into the cylinder, so we convert it to \(\langle 0,0,v\rangle\). Then